# A group of order $p^nm$ isn't simple

Suppose $$|G|=p^nm$$ with $$1. Show that $$G$$ is not simple.

I know it has to do with group actions. My idea is to consider a subgroup of order $$p^n$$, call it $$H$$. It has index $$m$$, so there are $$m$$ left cosets of $$H$$ in $$G$$, and $$S_m$$ acts on the set of left cosets. This gives a group homo $$G\to S_m$$. How do I proceed? I guess either $$H$$ should be normal or I should prove that the kernel of this map is a proper nontrivial subgroup, which is normal. But I don't know how to do either of those.

Sylow's theorem is the easiest way to go here. The number of $$p$$-Sylow subgroups is congruent to $$1\pmod p$$ and also divides $$m$$. Since $$m there must be only one $$p$$-Sylow subgroup, which is therefore normal. Thus $$G$$ is not simple because it has a normal subgroup of order $$p^n$$.
Or, to finish your proof, it is true that $$G$$ acts on the cosets of $$H$$. So there is an homomorphism from $$G$$ to $$S_m$$ which cannot be an isomorphism since $$p$$ does not divide $$m!$$, which is the order of $$S_m$$. Also, it can't map all of $$G$$ to $$1\in S_m$$ since the action on cosets is transitive. Thus, the kernel is a non-trivial normal subgroup, and $$G$$ is not simple.
You can use your idea to finish the proof. The kernel of the map is nontrivial, since $$p$$ does not divide the order of $$S_m$$.
Or you could use Sylow's theorems to prove that $$H$$ is normal.